1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <algorithm> 5 #include <map> 6 using namespace std; 7 const int maxn = 200005; 8 typedef long long ll; 9 struct car1 { 10 char number[10],state[10]; 11 int length; 12 }c1[10005],c2[10005]; 13 14 bool cmp1(car1 p,car1 q) { 15 if(strcmp(p.number,q.number) != 0) return strcmp(p.number,q.number) < 0; 16 else if(p.length != q.length) return p.length < q.length; 17 return strcmp(p.state,q.state) < 0; 18 } 19 bool cmp2(car1 p,car1 q) { 20 return p.length < q.length; 21 } 22 int query[90005];//存储查询时间 23 map<string,int>mp; 24 int main() { 25 int n,mm; 26 int k = 0;//有效记录数 27 int h,m,s; 28 int maxtime = 0;//停留的最长时间 29 scanf("%d%d",&n,&mm); 30 for(int i = 0; i < n; i++) { 31 scanf("%s %d:%d:%d %s",c1[i].number,&h,&m,&s,c1[i].state); 32 c1[i].length = h*3600 + m*60 + s;//换算成秒 容易比较 33 } 34 for(int i = 0; i < mm; i++) { 35 scanf("%d:%d:%d",&h,&m,&s);//将查询时间也换成秒 36 query[i] = h*3600 + m*60 +s; 37 } 38 sort(c1,c1+n,cmp1); 39 for(int i = 0; i < n-1; i++) { 40 if(strcmp(c1[i].number,c1[i+1].number) == 0 && strcmp(c1[i].state,"in") == 0 && strcmp(c1[i+1].state,"out") == 0) {//如果匹配 41 strcpy(c2[k].number,c1[i].number); 42 c2[k].length = c1[i].length; 43 strcpy(c2[k++].state,c1[i].state); 44 strcpy(c2[k].number,c1[i+1].number); 45 c2[k].length = c1[i+1].length; 46 strcpy(c2[k++].state,c1[i+1].state); 47 mp[c1[i].number] += c1[i+1].length - c1[i].length; 48 maxtime = max(maxtime,mp[c1[i].number]);//找到在停车场最长的停车时间 49 } 50 } 51 sort(c2,c2+k,cmp2);//将有效记录的车的时间按从小到大排序 52 int now = 0; 53 int car = 0; 54 //复杂度变成了O(n+m) ！！！ 55 //已知查询时间是从小到大的 56 for(int i = 0; i < mm; i++) {//关键 用now存储不大于该查询时间的车辆的时间 57 while(c2[now].length <= query[i]&& now < k) {//k是有效记录数 now一定要小于k 否则段错误！！ 58 if(string(c2[now].state) == "in") car++; //如果车辆的状态是in 则停车场中车数++ 59 else car--;//否则车数-- 60 now++;//判断下一辆车 61 } 62 printf("%d ",car); 63 } 64 map<string,int>::iterator it; 65 for(it = mp.begin(); it != mp.end(); it++) { 66 if(it->second == maxtime) printf("%s ",it->first.c_str()); 67 } 68 h = maxtime / 3600; 69 m = maxtime % 3600 /60; 70 s = maxtime % 3600 % 60; 71 printf("%02d:%02d:%02d",h,m,s); 72 return 0; 73 }